Unformatted text preview: Notes on Behaviour at Inﬁnity
The behaviour of a function y (t) is determined by the limit of y (t) as t → ∞.
Elementary Functions
The basic functions we need are powers of t and exponentials of t. These have the following limits.
lim tα = t→∞ ∞ if α > 0,
0
if α < 0 lim eαt = t→∞ ∞ if α > 0,
0 if α < 0 Of course if these functions are multiplied by a function, the sign of the limit can change: 3t2 → ∞ but
−5t7 → −∞.
Sums of Functions
Suppose we have y (t) = y1 (t) + y2 (t). Then in general we have
lim y (t) = lim y1 (t) + lim y2 (t), t→∞ t→∞ t→∞ and even when the limits are inﬁnite we can use the rules that for any real number a, ∞ + a = ∞ and
−∞ + a = −∞, together with the rules that ∞ + ∞ = ∞ and −∞ + (−∞) = −∞.
A problem occurs when we get ∞ + (−∞) which is not deﬁned. In this case we need to compare the
functions and determine which function grows to inﬁnity faster and is therefore the dominant term. If y1 (t)
dominates y2 (t) for large t, we express this by writing y1 (t)
y2 (t) as t → ∞. And in this case
lim y (t) = lim y1 (t), t→∞ t→∞ so the behaviour of y (t) is determined by the behaviour of the dominant term (and we can ignore the other
terms).
For exponentials and polynomials we have the follows dominance relations.
• For α, β > 0, eαt tβ as t → ∞. • For α > β > 0, eαt eβt as t → ∞. • For α > β > 0, tα tβ as t → ∞. Examples
lim −et + 11t−4 = −∞ t→∞ lim t2 + t−2 = +∞ t→∞ lim 1 + t−1 + t−2 + e−t = 1 t→∞ lim 3et − t2 − t7 = lim 3et = +∞ t→∞ t→∞ lim 5et − e2t = lim −e2t = −∞ t→∞ t→∞ lim t3 − t2 − t + 7 − e−8t
t→∞ +∞ 4t
2t
lim Ce − 3e = −∞
t→∞ −∞ = lim t3 = +∞
t→∞ if C > 0,
if C = 0,
if C < 0 +∞ if y0 > −2, t
−2
lim (y0 + 2)e + t + 3 = 3
if y0 = −2,
t→∞ −∞ if y0 < −2
lim (y0 + 2)e−2t + 4 = 4 t→∞ 1 ...
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 Spring '10
 ChristinaSaleh
 Limits, lim, t→∞

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